Electro electrician problems, electro electronic problems

The three-phase four-wire is 40 kilowatts, and the phase current is more than 80 amperes. The current carrying capacity is considered to be 5 ampere square centimeters, and 16 square copper wires can be used.

There are many standards for wiring, current carrying capacity, voltage drop, mechanical strength, economic current carrying capacity, if the current carrying capacity is equipped with 16 square meters.

Use 10 square copper wire or 16 square aluminum wire. The two upstairs were talking about the ampacity of the aluminum wire.

Summary. Hello, what are the specific questions?

Hello, what are the specific questions?

You can see it clearly.

The first question is voltage feedback.

Because it is not in the same phase, it is a voltage parallel feedback.

Yes, can you do two more?

It needs to be restructured.

That one. This side is only responsible for 1 or 2 questions.

Thank you.

The most typical non-inverting amplifier is a large multiple of voltage amplification (voltage gain coefficient).

Because of the grip of this. <>

Don't ask us how the conclusion refers to the leather socks, you need to ask your teacher, or you can read the textbook and study it yourself.

According to the imaginary shortness principle, u+=u-

According to the principle of virtual disconnection, it can be seen that there is no current on R2, and it can be inferred that u-=u+=ui is also based on the principle of virtual disconnection of the world, R1 and Rf are searched in series, and the current on them is i, and the equation can be established.

i=(uo-ui)/rf

i=(ui-0)/r1

UO=(1+RF R1)UI can be solved

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<> will give you a bright test of dismantling the world;

Full of spine Yiha.

The current of i is 2A, and according to Ohm's law, the voltage at both ends of R1 can be found to be 24V, the current flowing through R1 is 8A, the current of R3 is 4A, and the current of R2 and R4 is 6A respectively, so it can be seen that the current flowing through the line segment of I is 2A

Load Current:

i＝p/u＝40/110≈

Apparent power: S ui 220

Total Load Impedance:

z＝u/i＝220/

Resistance value: r u i 110

1》Inductive reactance: XL (z z r r) 604 604 302 302) 523 ( ).

2》Power Factor:

cosφ＝p/s＝40/80＝

If the power factor is increased to, the apparent power:

s1＝p/cosφ＝40/

Reactive power at power factor:

Q1 root number (s s p p) 80 80 40 40) 69 (var).

Reactive power at power factor:

Q2 root number (s1 s1 p p).

Reactive power compensation required:

q q1 q2 69 19 50(var)3》should be connected in parallel:

c＝q/(u×u×2×π×f×

Because the voltage obtained by adding z is a U-line. is a multiple of the z of the star join. The current to which z is added is i = u line z = phase z(Originally, the current was 10=u phase z).

The current of the ammeter is the combination of two currents on the switch.

i=10+(

is acting alone, us is considered a short circuit, since r2=r1, then the total current i2' = 1 2is=

US acts alone, IS is regarded as an open circuit, and the circuit resistance is left with R1 and R2 in series, i2''=US (R1+R2)=

So the branch current i2=i2'+i2''=